ON TRIANGULAR AND BAKERfS MAPS WITH GOLDEN MEAN AS

ON TRIANGULAR AND BAKERfS MAPS WITH GOLDEN MEAN AS THE PARAMETER VALUE Chyi-Lung Lin Department of Physics, Soochow University, Taipei, Taiwan 111, R.O.C. (Submitted January 1995)

1. INTRODUCTION We discuss the triangular map and also Baker's map [2], [4] with the parameter ju chosen with the value of the golden mean: (1 + V5)/2 « 1.618. For an arbitrary parameter value in the range 0 < ju < 2, the starting graph (x, /(*)) in the range 0 < x < 1 is a line segment for a triangular map, and two line segments for Baker's map (see Fig. 1 and Fig. 4). We are interested in the graph of / M , where n>\. Since the starting graph contains a set of line segments, the proceeding graphs ( x , / M ( x ) ) are then obtained from iterates of the beginning line segments in the starting graph. It is therefore important to discuss iterating these two maps on line segments. Since these two maps are piecewise linear maps, it can then be shown that the graph of / w is a composition of line segments (see Fig. 3 and Fig. 6). These two maps are simple for p in the range 0 < ju < 1 because the number of line segments does not increase under the action of mapping; the graph of / w is therefore simple. Yet, they are often complicated in the range ju>l, as the action of mapping on starting line segments will generate more line segments of which the lengths In general are different. The graph of / w Is then a set of line segments with Irregular shape which becomes very complicated when n is large. It is then difficult to determine the graph of / M . However, we can show that when // is chosen with specific values, for instance, the golden mean, the graphs (x, f^n\x)) are again simple. There are only a few types of line segments in each graph and, Interestingly, the numbers of line segments of the graphs are those of the Fibonacci numbers. Nature shows that Fibonacci numbers occur quite frequently In various areas; therefore, it is interesting to know that Fibonacci numbers and, In fact, Fibonacci numbers of degree m, can be generated from a simple dynamical system [3]. In this work, we contain some reviews of [3], show the similarity of these two maps when a specific parameter value Is chosen, derive geometrically a well-known identity In Fibonacci numbers, and show that some invariant sequences can be obtained. 2. THE TRIANGULAR MAP WITH /i = (1 + S) / 2 First, we discuss the general triangular T^ map, which is defined by TM(x) = l-ju\xl

(1)

xn+l = l-M\xnl

(2)

or where ju Is the parameter. We restrict the ranges to: -1 < x < 1 and 0 < // < 2, so that T^ maps from the interval [-1,1] to [-1,1]. Figure 1 shows three graphs of T^ for, respectively, // = 0.6,1, 1996]

423

ON TRIANGULAR AND BAKER'S MAPS WITH GOLDEN MEAN AS THE PARAMETER VALUE

and (1 + V5)/2. We define xx = TM(x) as the first iterate of x for TM, and xn = T^x^) = TW(x) as the /2th iterate of x for T^. Since all the graphs (x, Tjf](x)) are symmetrical for x>0 and x < 0 , we henceforth consider these graphs in only the region of x > 0 . The starting graph, ^(x) = 1 - / # , in the range x > 0 is then a line segment from point (0,1) to point (1,1 - ju) with slope -ju . We call this the starting line segment. Iterating this map on this starting line segment then generates all the proceeding Tj^ graphs. T (x)

(a) T (x)

\

0.8

/x = 1

0". 6 0. 4

0.2 0

(b) T (z)

0.75

\

/*=(l+iTF-)/2

C.2 5

-0.25

1-A, -0.75

(c)

FIGURE 1 424

[NOV.

ON TRIANGULAR AND BAKER'S MAPS WITH GOLDEN MEAN AS THE PARAMETER VALUE

For a general discussion on an iterate of a line segment, we consider an arbitrary line segment described by y = g{x) = a + bx, where 0 < x < 1 and -1 < g(x) < 1. The upper piece of this line segment, for which g(x) > 0, after one iteration, goes to a line segment described by y=TM(g(x)) = l~aju~ bjux. Hence, the slope and the length of the upper piece have been rescaled, the sign of the slope alters as well; the slope rescaling factor is seen to be -ju. The lower piece, for which g(x) S+L_. The graphs of Figure 2(b), 2(c), and 2(d) show an Iterate of a line segment of type L_, S+, and S_, respectively. From these, we conclude that the action of this map on line segments of these four types Is described as: T^L+^S+LS + ->L_ S_^L+ TM then acts as a discrete map for L and S. If we are only to count the number of line segments in a graph—we need not distinguish L + from L_, S+ from S_, and LS from SL—then (3) can be expressed more simply as: T,:L-^LS S->L From (4), we have the following results. Proposition 2.2: The graph of y - T^\x)

in the range 0 < x < 1 contains line segments of only

two types, type L and type S, and the total number of these line segments is Fn+V Proof: Since the starting graph is a line segment connecting points (0,1) and (1,1 - ju) [see Fig. 1(c)], it is thus a line segment of type L. From (4), we see that, starting from an L, line segments generated from iterates of that are, therefore, only of two types: type L and type S. From (4), we also see that line segments of type L and S, respectively, are similar to those rabbits of type large and small in the original Fibonacci problem; hence, the numbers of line segments of all the graphs Tjf] would be those of the Fibonacci numbers. Therefore, we have shown an interpretation of the Fibonacci sequence from the point of view of a simple iterated map. Although we start from a functional map T^ on a finite interval of x, however, if we take line segments as the entities, then TM acts as a discrete map for these entities, and the mechanism of generating line segments from the action of this discrete map is now the same as the breeding of the Fibonacci rabbits. We now let L(n) and S(«) represent, respectively, the numbers of L's and S's in the graph of T[f]. Then, from (4), we have h(n) = L(n-l) S(w) = L ( / i - l ) .

+

S(n-ll U

Equation (5) shows that L(w) = L(n - 1 ) + L(w - 2) and S(w) = S(n - 1 ) + S(/i - 2). Thus, both sequences {L(^)} and {$(n)} are the Fibonacci-type sequences. Since we start from an L with 1996]

427

ON TRIANGULAR AND BAKER'S MAPS WITH GOLDEN MEAN AS THE PARAMETER VALUE

slope - / / , we have L(l) = 1 and S(l) = 0. According to (5), we then have L(n) = Fn and $(ri) Fn_x, where Fn is the 72th Fibonacci number, and the |slope| of each line segment in the graph of T[f] is jun. Therefore, the total number of line segments in the graph of Tjf] is L(n) + S(n) = Fn +Fn__x - Fn+l. Figure 3 shows the graph of T*5], from which we can count the number of line segments as being F6 = 8.

FIGURE 3 In [1] there is an interesting theorem stating that limw_»oologC(>f, ju)/n = log ju, where C(n, ju) is the number of line segments of the graph y - f[n](x, ju) and ju is an arbitrary parameter value. In our case, C(n,ju) = Fn+l and // = (l + V5)/2, so using the well-known formula Fn = (//"-(1 -//)")/V5 we can easily calculate that X\mn^\ogC{n,ju)In is indeed the value log//. Proposition 2.3: A simple identity, juFn + Fn_x = jun. Proof: In the graph of T*?\ we have F„ long line segments and Fn_l short line segments, and the |slope| of each line segment is ff. We denote by d(L,n) andrf(S, n), respectively, the projection length of a line segment of type L and type S on the x-axis in the graph of T^"]. We then have d(L,ri)= (ju)l~n and d(S, n) = (//)"". Since the total projection length of these Fn+l line segments in the x-axis should be 1, we have Frid(L,n) + Fn_1d(S,n) = l

(6)

or ftF„ + Fn_1 = ff.

(7)

This is a well-known identity in Fibonacci numbers; here we have derived it geometrically. Proposition 2.4: There are three infinite LS sequences that are invariant under three iterations of the map (3). Proof: We consider the shape of the graph of y = Tjf](x) in the range 0 < x < 1. Since all the shapes are from the connection of line segments of type L+, L_, S+, and S_; therefore, we can describe each shape in terms of an LS sequence. We let LS[T"] represent the LS sequence 428

[NOV.

ON TRIANGULAR AND BAKER'S MAPS WITH GOLDEN MEAN AS THE PARAMETER VALUE

describing the shape of the graph of y = Tjj"](x) in the range 0 < x < 1. The starting graph is simply L_. From (3), we have: LS[T!] = L_ LS[T2] = L+S_ LS[T3] = S+L_L+ LS[T4] = L_L+S_S+L_ LS[T5] = L+S_S+L_L+L_L+S_ LS[T6] = S+L_L+L_L+S_S+L_L+S_S+L_L+

,g)

LS[T"] = LS[r- 3 ]LS[r- 2 ]LS[T"- 4 ]LS[T"- 3 ] The last formula in (8) is for general n and can be proved easily by induction: starting from « = 5, we have LS[T5] = LS[T2]LS[T3]LS[T1]LS[T2], and then, after one iteration, we have LS[T6] = LS[T3]LS[T4]LS[T2]LS[T3], ... etc. The length of the LS[T"] sequence is Fn+l. Equation (8) shows that, after three iterations, an LS sequence will get longer but the original LS sequence remains. By taking n -> oo, we then have an infinite LS sequence which is invariant after three iterations of the map (3), since from (8) we have limLS[r]=limLS[T"- 3 ]. »-»oo

»->oo

We denote by {T*} the first Invariant Infinite LS sequence obtained from iterates of an L_, that is, {T7°}=limT*3w](L_). We see that (I™} is invariant after three iterations of the map (S)rThere are two other invariant infinite sequences which we denote by {T£°} and {TJ0}, where {TJ0} is obtained from an iterate of {IT}, i-e., {T2W} = T^Tr), and {%"} is from an iterate of {T~}, i.e., {%>} = T ^ } = T ^ T f } . Therefore, there are at least three infinite LS sequences that are invariant after three iterations of the map (3). Since {TH = T {TH = lim T^+l\h.)

= lim T^(L + S_) = lim Xf^(L + ),

this means that {T2°°} can be obtained from iterates of an L+. Finally, since {T3-} = T {T2°°} = lim Tf+1^(L+) = lim Tf^(S+L_) = lim TJ^(S + ), this means that {T3°°} can be obtained from iterates of an S+. Therefore, from the first few iterates of L+, L_, and S+, we have the following first few elements of these three infinite sequences: \\-\

j — 1^_ JL+o_ 0+L—JL+o—o+JL—JL+JL—JL+O_o-j-L'—l_y+JL—JU+O—o+JL'— . . .

{T"} = L+S_S+L_L+L_L+S_S+L_L+L_L+S_S+L_L+S_S+L_L+ ...

(9)

{T3"} = S+L_L+L_L+S_S+L_L+S_S+L_L+L_L+S_S+L_L+S_S+ ... There is no invariant sequence with S_ as the first element since, after three iterations, S_ goes to L_LH.S_, the first element is now L_ instead of S_. If a sequence were an Invariant sequence, its 1996]

429

ON TRIANGULAR AND BAKER'S MAPS WITH GOLDEN MEAN AS THE PARAMETER VALUE

length must be infinite and its first element can only be an L+ or an L_ or an S+, and that would just correspond to the infinite sequences 07°}, {T2°°}, and {T3°°}; hence, there are only three infinite LS sequences that are invariant after three iterations of the map (3). As a result, to arrange LS sequences that are invariant after three iterations of the map (3), the L+'s, L_'s, S+'s, and S_!s should be arranged according to the orders described in (9). This interesting phenomenon may have applications in genetics. The general rule for deciding the /1th entry in each of these three sequences is complicated; we will discuss this in the next section, where we treat a similar but simpler case. 3. BAKER1 S MAP WITH /i = (1 + V5) / 2 We now consider the easier Baker B map, which is defined by B„(*) =

fJX

//(JC-1/2)

for0 oo, we then have an infinite LS sequence which is invariant under the action of map (11) since, from (12), we have limLS[B"]=limLS[B"- 1 ]. W-»00

W-»00

We denote by {B°°} the infinite LS sequence obtained from iterates of an L, that is, {B°°}=limB[;](L). We see that {B00} is invariant under the action of map (11). It is impossible to have an invariant sequence with S as the first element. Therefore, we have only one infinite LS sequence that is invariant under the action of map (11). Therefore, from the first few iterates of L, we have the following first few elements of this infinite sequence: {B00} = LSLLSLSLLSLLSLSLLSLSL ...

(13)

Thus, to have an invariant LS sequence, the L's and S's should be arranged according to the order described in (13). We shall discuss this special symbol sequence in more detail. Denoting by B(k) the k^ element of this sequence, then, from (13), we have B(k) = L for k = 1, 3, 4, 6, 8, 9, 11, 12, 14, ..., and B(k)= S for k = 2, 5, 7, 10, 13, 15, ... . It would be interesting to find a way 432

[NOV.

ON TRIANGULAR AND BAKER'S MAPS WITH GOLDEN MEAN AS THE PARAMETER VALUE

of determining that B(k) is an L or an S for a given k. So far, we have not obtained a simple formula for this, except for the following descriptions that are based on the following theorem. Theorem: For k > 3 , B(k) = B(d), where k = d + Fn(k) mdFn(k) is the largest Fibonacci number that is less than k. Proof: Suppose Fn^ is the largest Fibonacci number that is less than k, and let d - k - Fn^. Then Fn^